If $A$ is a square matrix of order $3 \times 3$,then $|\operatorname{adj} A| =$ . . . . . . .

Let $k$ be a positive real number and let $A = \begin{bmatrix} 2k-1 & 2\sqrt{k} & 2\sqrt{k} \\ 2\sqrt{k} & 1 & -2k \\ -2\sqrt{k} & 2k & -1 \end{bmatrix}$ and $B = \begin{bmatrix} 0 & 2k-1 & \sqrt{k} \\ 1-2k & 0 & 2\sqrt{k} \\ -\sqrt{k} & -2\sqrt{k} & 0 \end{bmatrix}$. If $\det(\operatorname{adj} A) + \det(\operatorname{adj} B) = 10^6$,then $[k]$ is equal to [Note: $\operatorname{adj} M$ denotes the adjoint of a square matrix $M$ and $[k]$ denotes the greatest integer less than or equal to $k$].

If $A$ is a square matrix of order $3$, then $|\operatorname{Adj}(\operatorname{Adj} A^2)|=$

If $A = \left[\begin{array}{cc}1+2 i & i \\ -i & 1-2 i\end{array}\right]$ where $i=\sqrt{-1}$,then $A (\operatorname{adj} A )=\ldots$. (in $I$)

If $A=\left[\begin{array}{ll}2 & 3 \\ 1 & 2\end{array}\right]$ and $B=\left[\begin{array}{cc}2 & -3 \\ -1 & 2\end{array}\right]$,then $\left(B^{-1} A^{-1}\right)^{-1}=$

If $A=\left[\begin{array}{ccc}1 & 2 & 1 \\ -1 & 1 & 3\end{array}\right]$ and $B=\left[\begin{array}{cc}1 & 2 \\ -3 & 1 \\ 0 & 2\end{array}\right]$,then find $(AB)^{-1}$.